132_ch16 - PHYS 132 WEEK 7 CH 16 16.9 Solve Because the...

PHYS 132, WEEK 7, CH 16
16.9.
Solve: Because the atomic mass number of gold is 197, 1.0 mole of gold has a mass of 197 g or 0.197 kg. The volume of 1.0 mole
of gold is
V = L
3
=
M
r
Au
L =
0.197 kg
( )
19, 300 kg m 3
1 3
= 2.17 cm
16.12.
Solve: Let T
F
= T
C
= T :
T
F
= 9
5
T
C
+ 32 ° T = 9
5
T + 32 ° T = - 40 °
That is, the Fahrenheit and the Celsius scales give the same numerical value at - 40 ° .
16.19.
Model: Treat the gas in the sealed container as an ideal gas.
Solve: (a) From the ideal gas law equation pV = nRT , the volume V of the container is
V =
nRT
p
=
2.0 mol
( )
8.31 J mol K
( )
273 + 30
( )
K
[ ]
1.013 10 5 Pa
= 0.0497 m
3
Note that pressure must be in Pa in the ideal gas law.
(b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is
p
1
V
T
1
=
p
2
V
T
2
p
2
= p
1
T
2
T
1
= 1.0 atm
( )
273 + 130
( )
K
273 + 30
( )
K
= 1.33 atm
Note that gas-law calculations must use T in kelvins.
16.24.
Model: Treat the oxygen gas in the cylinder as an ideal gas.
Solve: (a) The number of moles of oxygen is
n =
M
M
mol
=
50 g
32 g mol
= 1.563 mol
(b) The number of molecules is N = nN
A
= 1.563 mol
( )
6.02 10
23
mol
- 1
( )
= 9.41 10
23
.
(c) The volume of the cylinder
V = p r
2
L = p 0.10 m
( ) 2
0.40 m
( )
= 1.257 10
- 2
m
3
.